Question

A movie with an aspect ratio of 1.25:1 is shown as a pillarboxed image on a 36-inch 4:3 television. Calculate the Areas of the TV, the Image and One Blackbar

Answer 1

Step-by-step explanation

The television has a diagonal that measures 36 inches:

And the ratio is 4:3

`\begin{gathered} (w)/(h)=(4)/(3) \\ w=(4)/(3)h \end{gathered}`

We can use the Pythagorean theorem to find the height of the TV:

`\begin{gathered} 36^2=h^2+w^2 \\ 36^2=h^2+((4)/(3)h)^2 \\ 36^2=h^2(1+(4^2)/(3^(2))) \\ 1296=h^2(1+(16)/(9)) \\ 1296=h^2*(25)/(9) \\ h^2=1296*(9)/(25) \\ h=\sqrt[]{1296*(9)/(25)}=21.6 \end{gathered}`

The height of the TV is 60 inches. It's width is:

`w=(4)/(3)h=(4)/(3)*21.6=28.8`

w=80 inches

Therefore the area of the TV is

`A_(TV)=w* h=28.8*21.6=622.08in^2`

The move has an aspec ratio of 25:1 shown as a pillarboxed image. This means that this is what we see:

So we know that the image height is the same as the TV's, 21.6 inches.

The relation between it's height and it's width is:

`\begin{gathered} (w)/(h)=(1.25)/(1) \\ w=1.25h \\ \text{if h = 21.6 in} \\ w=27in \end{gathered}`

The area of the image is:

`A_{\text{image}}=w_{\text{image}}* h=27*21.6=583.2`

The area of the two blackbars is the difference between the area of the TV and the area of the image:

`A_(2-blackbars)=A_(TV)-A_(image)=622.08-583.2=38.88in^(2)`

Since we need to find the area of just one blackbar, we just have to divide the area of both blackbars by 2:

`A_(1-blackbar)=(A_(2-blackbars))/(2)=(38.88)/(2)=19.44in^(2)`

Answer

• Area of the TV: ,622.08 in²

,

• Area of the image: ,583.2 in²

,

• Area of one blackbar: ,19.44 in²